Question

The jib crane is supported by a pin at Cand rod AB. The rod can withstand a maximum tension of 40 kN. If the load has a mass of 2000 kg, with its center of mass located at G. Determine its maximum allowable distance x and the corresponding horizontal and vertical components of reaction at C.

Answer 1

Answer:

The maximum allowable distance = 5 m

Explanation:

Data:

There will be three forces on the jib. Let the forces be denoted as:

$C_{x}$, $C_{y}$ and the force on pole AB

To find the angle AB makes with the horizontal beam:

$tan^{-1}(\frac{3}{4}) = 36.8699$

The load has a mass of 2 000 kg then, the force will be:

F = mg, where g = 9.81 m/s²

  = $2000* 9.81 = 19 620 N$

Breaking AB into its x and y coordinates:

$AB_{x}=ABcos(36.8699)\\AB_{y} = ABsin(36.8699)$

Then,

∑$M_{c}$ = 0

$0 = (4)(AB sin (36.8699) + 0.2 (AB cos (36.8699) - (5*19620)\\AB = 38 320. 3 N$

∑$F_{x} = 0\\C_{x} = AB cos(36.8699)\\C_{x} = 30 656.2 N$

∑$F_{y} = 0\\C_{y} + AB sin (36.8699) - 19 620 = 0\\C_{y} = - 3 372.18 N$

so the components of the forces will be 30 656.2 N and - 3 372.18 N